Question

The 4th and the last terms of an A.P. are 11 and 89 respectively. If there are 30 terms in the A.P., find the A.P. and its 23rd term.​

Answer 1

$\underline{\underline{\large\bf{Given:-}}}$

$\red{\leadsto}\:$$\textsf{}$$\sf Number \: of \:terms \: in \: A.P,n = 30$

$\red{\leadsto}\:$$\textsf{}$$\sf Fourth \: term ,a_4 = 11$

$\red{\leadsto}\:$$\textsf{}$$\sf last\:term, a_{30} = 89$

$\underline{\underline{\large\bf{To Find:-}}}$

$\orange{\leadsto}\:$$\textsf{ }$$\sf The \: A.P.$

$\orange{\leadsto}\:$$\textsf{ }$$\sf 23rd\: term, a_{23}$

$\underline{\underline{\large\bf{Solution:-}}}$

The nth term of A.P is determined by the formula-

$\green{ \underline { \boxed{ \sf{a_n = a+(n-1)d}}}}$

where

• $\sf a = first \:term$
• $\sf a_n = nth \: term$
• $\sf n = number \:of \:terms$
• $\sf d = common \: difference$

Since ,

$\sf a_4 = 11$

$\longrightarrow$ $\sf a+(4-1) d= 11$

$\longrightarrow$ $\sf a+3d= 11\_\_\_(1)$

$\sf a_{30}= 89$

$\longrightarrow$ $\sf a+(30-1)d=89$

$\longrightarrow$$\sf a+29d= 89\_\_\_(2)$

Subtracting equation (1) from equation(2)

$\begin{gathered}\\\implies\quad \sf a+29d-(a+3d) = 89-11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a+29d-a-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a-a+29d-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf 26d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = \frac{78}{26} \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = 3 \\\end{gathered}$

Putting the value of d in equation (1) -

$\begin{gathered}\\\implies\quad \sf a+3(3) = 11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 11-9 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 2 \\\end{gathered}$

• First term of A.P, a = 2

• Second term of A.P.,$\sf a_2= 2+(2-1)\times 3$

$\quad\quad\quad\sf =2+3$

$\quad\quad\quad\sf =5$

• Third term of A.P.,$\sf a_3= 2+(3-1)\times 3$

$\quad\quad\quad\sf =2+6$

$\quad\quad\quad\sf =8$

$\longrightarrow$Thus , The A.P is 2,5,8,. . . . . .

Now,

$\begin{gathered}\\\implies\quad \sf a_n = a+(n-1)d \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23 }= 2+(23-1)\times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+22 \times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+66 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 68 \\\end{gathered}$

$\longrightarrow$Thus , 23rd term is 68.